# Rescaling time in differential equations

On a scientific paper, I found the following equations about a compass gait (one leg behaves like an inverted pendulum, the other one as a simple pendulum; $$\theta$$ and $$\phi$$ are time-dependent):

$$\ddot{\theta} - \sin\left( \theta - \gamma \right) = 0$$

$$\ddot{\theta} - \ddot{\phi} + \dot{\theta}^2 \, \sin(\phi) - \cos(\theta - \gamma) \, \sin(\phi) = 0$$

I got the following ones, and also, in the paper, there are these before the sentence (*):

$$\ddot{\theta} - \frac{g}{l} \, \sin\left( \theta - \gamma \right) = 0$$

$$\ddot{\theta} - \ddot{\phi} + \dot{\theta}^2 \, \sin(\phi) - \frac{g}{l} \, \cos(\theta - \gamma) \, \sin(\phi) = 0$$

(*) The explanation of the paper to pass from the last two equations to the first two equations is "we have rescaled time by $$\sqrt{l/g}$$".

Could you tell me what it means?

• If you remove the last part "how to get the first two equations from the last two" then it will be on-topic. Commented May 15 at 14:25
• Which scientific paper? Which page? Commented May 15 at 22:25
• @Qmechanic pag. 2 of Mariano Garcia paper: The Simplest Walking Model: Stability, Complexity, and Scaling Commented May 16 at 8:51
• Permalink: doi.org/10.1115/1.2798313 Commented May 16 at 8:55

It's an usual procedure in deriving non-dimensional equations, from the dimensional ones: angles have no physical dimensions, they wanted a "scaled" (non-dimensional) time as well.

You just need to define a "scaled-time" independent variable

$$\tau = \sqrt{\frac{g}{\ell}} t$$

perform derivatives using the rules for composite functions,

$$\phi'(\tau) := \frac{d}{d\tau} \phi(t(\tau)) = \frac{dt}{d\tau} \frac{d}{dt} \phi(t) = \sqrt{\frac{\ell}{g}} \dfrac{d \phi}{d t} = \sqrt{\frac{\ell}{g}} \dot{\phi}(t) \ ,$$

$$\rightarrow \qquad \phi'' = \frac{\ell}{g} \ddot{\phi} \quad , \quad \phi'^2 = \frac{\ell}{g} \dot{\phi}^2$$

and multiply all the equations by the factor $$\frac{g}{\ell}.$$

• You could add that if $g$ and $l$ mean what they usually mean, indeed $\sqrt{g/l}$ has dimension of $1/\mathrm{time}$, so $\tau$ is dimensionless Commented May 15 at 10:08
• Yeah, I was assuming that. Even if they don't mean what they usually mean, it's enough that $\theta$ is dimensionless (like angles are) and time $t$ has the proper dimension in the original equations: this is enough to imply that $\sqrt{g/l}$ has the physical dimension $1/\text{time}$ for the equations to be consistent Commented May 15 at 10:19
• Let's go, anonymous downvoter! Keep pushing! Commented May 15 at 13:30

It is the non-dimensionalization of the last two differential equations. Assuming $$g$$ as the acceleration due to gravity ($$\text{m}/\text{s}^2$$) and $$l$$ as the length (m), $$\sqrt{l/g}$$ has the dimension of time. Rescaling time by $$\sqrt{l/g}$$ means the substitution of the (physical quantity) time $$t$$ by its nondimensionalized counterpart $$\tau=t/\sqrt{l/g}$$.

Note that $$\tau=t\sqrt{\frac gl}$$ and $$\frac{\rm d\tau}{{\rm d}t}=\sqrt{\frac gl}$$.

I want to explain this in full detail since just writing $$\theta(\tau)$$, $$\phi(\tau)$$ and $$\gamma(\tau)$$ feels wrong (abuse of notation) and non-intuitive to me.

Let $$\vartheta(\tau):=\theta(t)=\theta(\tau\sqrt{l/g})$$, $$\varphi(\tau):=\phi(t)=\phi(\tau\sqrt{l/g})$$, and $$\eta(\tau):=\gamma(t)=\gamma(\tau\sqrt{l/g})$$.

Now, by the chain rule: $$\dot\theta:=\dot\theta(t)=\frac{{\rm d}\theta}{{\rm d}t} =\frac{{\rm d}\vartheta}{{\rm d}\tau} \frac{{\rm d}\tau}{{\rm d}t} =\sqrt{\frac{g}{l}} \ \dot \vartheta(\tau)$$ $$\ddot\theta:=\ddot\theta(t) =\frac{{\rm d}\dot\theta}{{\rm d}t} =\sqrt{\frac{g}{l}}\frac{{\rm d}\dot\vartheta}{{\rm d}\tau} \frac{{\rm d}\tau}{{\rm d}t} =\frac{g}{l}\ \ddot \vartheta(\tau)$$

Similarly, $$\dot\phi:=\dot\phi(t)=\sqrt{\frac{g}{l}} \ \dot \varphi(\tau),\ \ddot\phi:=\ddot\phi(t)=\frac{g}{l}\ \ddot \varphi(\tau)$$

Substituting these values into the last two differential equations, we get $$\color{red}{\boxed{\ddot{\theta} - \frac{g}{l} \, \sin\left( \theta - \gamma \right) = 0}}\implies\frac{g}{l}\ddot{\vartheta} - \frac{g}{l} \, \sin\left(\vartheta - \eta \right) = 0\implies\color{blue}{\boxed{\ddot{\vartheta} - \sin\left( \vartheta - \eta \right) = 0}}$$

$$\color{red}{\boxed{\ddot{\theta} - \ddot{\phi} + \dot{\theta}^2 \, \sin(\phi) - \frac{g}{l} \, \cos(\theta - \gamma) \, \sin(\phi) = 0}}\implies \frac{g}{l} \,\ddot{\vartheta} - \frac{g}{l} \,\ddot{\varphi} + \frac{g}{l} \,\dot{\vartheta}^2 \, \sin(\varphi) - \frac{g}{l} \,\cos(\vartheta - \eta) \, \sin(\varphi) = 0\implies\color{blue}{\boxed{\ddot{\vartheta} - \ddot{\varphi} + \dot{\vartheta}^2 \, \sin(\varphi) -\cos(\vartheta - \eta) \, \sin(\varphi) = 0}}$$ which are the required differential equations (in non-dimensionalized form) after rescaling time by $$\sqrt{l/g}$$.

I hope it helps!